Imagine one circle with radius 5, drop inside this parabola and "stuck there" (hence there is some area between the circle and parabola)

1) find the centre of this circle (should be easy)hint: obviously x coordinate is 0

Imagine another circle, with radius 4, drop inside this parabola on top of the previous circle. (so now there are 2 circles in the parabola) Unlike the first circle, this smaller circle is against the right hand side of the parabola.

2) Find the centre of this circle (challenging)

Sorry for my poor description, but just imagine you have a plastic bag (with shape y=x^2 fixed). You drop one ball into this bag, then drop another.

Does anyone know whether i can attach some picture files in this forum? So that anyone can view the graph clearly.

Yup, the 2nd question (created by myself) quite hard. In fact i totally spent more than 10 hrs on this question.

Actually the original question was finding the area between 2 circles and the parabola (the little bending triangle). But I guess the ppl who is able to find the small circle's centre is almost surely able to find the area.

Finally I managed to get an answer,though not an exact one, but a good approximation.....

I found the center of the circle with radius 4:x=1.82167.....y=34.06371....Here I will give a short description of my method.

1)First of all, the distance from the center of the first circle (0,25.25) to the center of the second circle is 9, so the equation of the center of the second circle is:y(x)= (81-x²)^½+25.25...............eq(1)

2)Now we have to find some point on this circle (eq(1)), this point should be the center the second circle, and we know the distance from this point , perpendicular to the tangent of the parabola, is 4.

In other words, we have to construct an intersecting point.I will find this intersecting point by finding another parabola ,parallel to the given one.

3)The second parabola I found this way:First of all I chose 3 points P,Q and R on the first parabola, with coordinates and slope:P:x_1=5,y_1=25 and y_1'=10Q:x_2=6,y_2=36 and y_2'=12R:x_3=7,y_3=49 and y_3'=14

So the slopes at these 3 points is known here, and also the slope of their normals.

Using this I found the coordinates of 3 points at a distance of 4 from P,Q and R.Now I can set up 3 equations with 3 unknowns:y(x)=Ax²+BX+C.

I think I figured out a way to do it. You are going to need a bunch of equations though.

Equation of Top CircleEquation of Bottom CircleDerivative of Top CircleDerivative of Bottom Circle

Now, we can set the two pairs of respective equations equal and solve somewhat. But, we still have four unknowns so we need 4 equations. My idea is that the length of the line segment connecting the two centers is going to be 9. Therefore, we can use the distance formula to produce a third equation. The fourth equation could possibly be produced by setting the opposite of the concavities equal (I don't think this is legal though b/c the circles are of different radii).

However, these equations quickly reach high exponents making them unsolvable by algebraic means.

As you mentioned, we do need a few equations to get the circle centre. At first I tried using this method and i got to solve an 8-degree polynomial equation (i.e. solve ax^8 + bx^7 + ... + hx + i = 0) something like that.

Then I tried to find some "more efficient" equations. I managed to reduce the degree of polynomial to 6. Then 4. Few weeks ago I managed to reduce it to 3-degree. Then I applied cardano's formula to solve.

When I kept on to try to reduce 2-degree (i.e. quadratic, which means that anyone above grade 10 is able to solve it), suddenly a bad news came to me. I found that there was a mistake in my calculation (the cubic equation), so far I havent found out the reason. (but i am sure that the error is a concept error, not calculation error)

Means that the best result I could get so far is to reduce the equation to 4-degree. I am going to have a big exam on May 7, after this day I will concentrate on finding out the mistake i made.

1)First of all, the distance from the center of the first circle (0,25.25) to the center of the second circle is 9, so the equation of the center of the second circle is:y(x)= (81-x²)^½+25.25...............eq(1)

This should read:1)First of all, the distance from the center of the first circle (0,25.25) to the center of the second circle is 9, so the equation of the circle on which the center of the second circle is located is:y(x)= (81-x²)^½+25.25...............eq(1)

This explains the need for a second curve,parallel to the first parabola, as described in 3).